Consider the problem of packing an upwards-pointing unit equilateral triangle "efficiently" by downwards-pointing equilateral triangles, where "efficiently" means that there is little wasted area relative to the perimeter of the triangles used in the packing. The $n^{th}$ generation of the Sierpinski triangle

packs all but $(3/4)^n$ of the area of the large upwards triangle by downwards triangles, at the cost of a net perimeter of $O((3/2)^n)$. Thus, if we let $\varepsilon$ denote the area not packed, then the perimeter of the triangles used in this construction is $\gg \varepsilon^{-\alpha}$ for $\alpha = \frac{\log (3/2)}{\log (4/3)} = 1.409\dots$.

My question is whether this phenomenon is general: given any finite collection of downward equilateral triangles in the upward unit equilateral triangle that is a packing (i.e., interiors are disjoint) and leaves an area of $\varepsilon$ not covered, is it true that the total perimeter of the triangles used is of the form $\gg \varepsilon^{-c}$ for some absolute constant $c>0$? For my application I do not need an optimal exponent $c$.

I think I can establish a bound of the form $\gg \log \frac{1}{\varepsilon}$ (roughly speaking, by arguing that every dyadic scale of triangles between $\varepsilon$ and $1$ has to contribute a constant amount of perimeter, otherwise there will be too much waste), but for my application I really need a polynomial lower bound (or maybe $\exp( (\log\log \frac{1}{\varepsilon})^C )$ for a large absolute constant $C$ might suffice). It's intuitively plausible to me that the Sierpinski packing is the "best" packing for this purpose, and that the smaller triangles really have to contribute more than a constant amount of perimeter, but I am finding it surprisingly tricky to locate a rigorous argument. Perhaps this sort of question has already been studied in the literature?